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EMSE-515-F05-08
EMSE-515
Fall 2005
F. Ernst
Rutherford Backscattering Spectrometry
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Bohr’s Model of an Atom
• existence of central core established by single collision,large-angle scattering of alpha particles (4He2+)
! basis of Rutherford back-scattering spectrometry (RBS)
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Principle of RBS
• exposed specimen surface to beam of (light) MeV particles
• elastic collisions with (heavy) atoms of target
! Coulomb scattering in a central-force field
• can be described by classical mechanics
• but: specimen will eventually stop beam particles
! su!cient penetration of the target requires beam particleswith kinetic energy in the MeV range
! accelerator!
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NEC 5SDH Accelerator
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NEC 5SDH Accelerator
• 1.7 MV tandem pelletron
• 3.4 MeV protons
• 5.1 MeV alpha particles
• N ions with energies in excess of 7.0 MeV
• detectors
! Si surface barrier detector
! NaI(Ti) scintillator
! liquid nitrogen-cooled Si(Li) detector
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NEC 5SDH Accelerator
• detectors enable detection of
! scattered ions
! characteristic gamma rays
! characteristic x-rays
• can be used for the following experimental techniques:
! RBS
! PIXE (particle induced X-ray emission)
! NRA (nuclear reaction analysis)
! sample temperatures from 77 to 1000 K
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Detection of Scattered Particles
• semiconductor nuclear particle detector
• resolve rate and kinetic energy of scattered particles
• most systems use a surface-barrier, solid state detector
! Schottky-barrier diode
! reverse-biased
! incident particle generates electron–hole pairs
! collected to electrodes
! voltage pulse, proportional to particle energy
! collect counts in voltage bins of multichannel analyzer
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Surface Barrier, Solid State Detector
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Surface Barrier, Solid State Detector
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Resulting Spectrum Properties
• statistics of electron-hole pair generation
• noise in output signal
• limited energy resolution
• typical energy resolution: 10 . . . 20 keV
• backscattering analysis with 2.0 MeV 4He particles can re-solve isotopes up to about mass 40 (clorine isotopes, forexample)
• mass resolution decreases with increasing atomic mass ofthe target
• example: 181Ta and 201Hg cannot be distinguished
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Experimental Setup for RBS
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Experimental Setup for RBS
• beam of mono-energetic particles (4He)
• in vacuum
• backscattering
! energy transfer to (stationary) target atoms
! primary particle looses kinetic energy
• amount depends on masses of incident particle and targetatom
• energy loss provides “signature” of target atoms
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Characteristics of RBS
• multi-element depth concentration profiles
• fast, non-destructive analysis (no sample preparation orsputtering required)
• matrix independent (una!ected by chemical bondingstates)
• quantitative without standards
• high precision (typically ±3 %
• high sensitivity
! e. g. 1011 Au/cm2 on Si
! depends on Z and sample composition
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Characteristics of RBS
• depth range: between 0 and 1 mm
• depth resolution !2 nm near the surface
• spatial definition:
" beam spot size 0.5 . . .2.0 mm
" map or raster option to 7 cm#7 cm
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Typical Applications of RBS
• absolute thickness of films, coatings, and surface layers(in atoms/cm2)
• surface/interface contaminant detection(oxide layers, adsorbates, etc.)
• interdi!usion kinetics of thin films (metals, silicides, etc.)
• elemental composition of complex materials(phase identification, alloy films, oxides, ceramics, etc.)
• quantitative dopant profiles in semiconductors
• process control monitoring
! composition
! contaminant control
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Kinematics of Elastic Collisions
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Kinematics of Elastic Collisions
• RBS: elastic Coulomb scattering
! calculate energy transfer in the framework of classical me-chanics
• applying the conservation principle of energy and momen-tum
• energy:M12v2 = M1
2v2
1 +M22v2
2
v: velocity of beam particle before collision;v1: velocity of beam particle after collision;v2: velocity of target particle after collision;M1: mass of beam particle;M2: mass of target particle.
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Kinematics of Elastic Collisions
• momentum:
p = p1 + p2
!! M1v = M1v1 +M2v2
p: momentum of beam particle before collision;p1: momentum of beam particle after collision;p2: momentum of target particle after collision.
• components parallel and normal to the axis (direction ofthe incident beam):
M1v = M1v1 Cos[!]+M2v2 Cos[!]
" 0 = M1v1 Sin[!]+M2v2 Sin[!]
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Kinematics of Elastic Collisions
• for the ratio of particle velocities, these expressions yield
v1v=±!M2
2 !M21 Sin[!]2 +M1 Cos[!]M1 +M2
,
where the plus sign holds for
M1 < M2
• ratio of the projectile energies for M1 < M2 is known as“kinematic factor” KM2
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Kinematic Factor
• KM for a scattering angle ! = 170! as a function of thetarget mass M2 for di!erent beam particles:
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Kinematic Factor
• the sensitivity for small di!erences !M2 becomes a maxi-mum for ! = 180!
" ! = 180! is the preferred position for the detector
# “back-scattering”
• in practice (to accommodate the detector size):detector at ! $ 180!
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Example of RBS Spectra
• !1 ML of di!erent noble metals:
" 63, 65Cu
" 107, 109Ag
" 197Au
" ! = 170"
• 2.5 MeV 4He ions
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Example of RBS Spectra
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Discussion
• RBS can detect less than one monolayer of heavy elements!
• peaks representing the elements
! well separated
! easy to identify
• determination of absolute coverage?
" requires knowledge on absolute cross-section (see below)
• limits of mass resolution # separation of isotope peaks
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Discussion
• Cu:
! 63Cu (K = 0.777) and 65Cu (K = 0.783)
! energy di!erence !E1 = 17 keV for 2.5 MeV 4
" somewhat larger than detector resolution (#15 keV)
$ Cu isotopes are resolved
• Ag:
! 107Ag and 109Ag
! !E1 = 6 keV
" smaller than energy resolution of detector
! Ag isotopes not resolved
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Scattering Cross Section
• probability of scattering is equal to fraction of sample area“blocked” by scatters
• important concept for scat-tering problems:
scattering cross section
! e!ective “ target area”presented by each scatter(atom)
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Scattering Cross Section
• scattering cross section can be considered for specifickinds of scattering events, for example:
! elastic scattering " !el
! inelastic scattering " !inel
• thin samples:
! target areas of individual scatters do not overlap
" N scatterers block area N · !! fraction of “rays” removed from incident beam:
N!A
= N!xAx
= "!x
x: sample thickness; " # N/Ax: number density of scatterers.
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Impact Parameter
• scattering is generally anisotropic
• depends on impact parameter b
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Di!erential Scattering Cross-Section
• scattering angle ! decreases with increasing b
• for increasing b, ! ! 0 (forward scattering)
• angular distribution of scattered particles?
• important concept:
di!erential scattering cross section:
d"d!
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View Along Incident Beam Direction
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Di!erential Scattering Cross Section
• interpretation of di!erential scattering cross section:
! piece of area d! o!ered by the scatterer for scatteringinto a particular increment d! in solid angle
! d! varies with the location of d!, characterized by thescattering angle ":
d!d!
= d!d!["]
• relation between increment in scattering angle " and incre-ment in solid angle:
d! = 2# Sin["]d"
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Di!erential Scattering Cross-Section
• relation between increment in scattering angle ! and incre-ment in solid angle:
d! = 2" Sin[!]d!
• in backscattering spectrometry, the detector solid angle!D is small – typically < 10!2 sterad
• therefore, one often uses an average di!erential cross-section:
#[!] = 1!D
!
!D
d#d!
d!,
where # is usually called “scattering cross-section”
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Di!erential Scattering Cross-Section
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Quantification
• consider thin target of thickness t with N atoms/cm3, thus
Ns = Nt
target atoms per unit area
• total number of incident particles: Q
• number of particles detected by the detector positioned atscattering angle !: QD[!]
! to infer Ns from QD, one needs to know "[!], the “scatter-ing cross-section:”
QD[!] = "[!] ·!D ·Q ·Ns
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Quantification
• !["] can be calculated from the force that acts during thecollision
• since high-energy particles penetrate up to the core ofthe target atom, this force mainly corresponds to an un-screened Coulomb repulsion between the two positivelycharged nuclei
• one-body formulation of this problem:
! scattering by a central force
! conservation of kinetic energy
! impact parameter b
! rotational symmetry
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Quantification
• particles with impact parameters between b and b+db aredeflected into an annular region spanned by scattering an-gles between ! and ! + d!
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Quantification
• the area of this region is
d! = 2! Sin["]d"
• recall definition of the scattering cross-section:
d# = #["]d!
! annular area enclosing scattering angles [", " + d"] is re-lated to annular area
d# = 2!bdb
by
2!bdb = "#["] · 2! Sin["]d"
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Central Force Scattering
• geometry of RBS:
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Central Force Scattering
• central force!
! scattered particle travels on a hyperbolic trajectory
• elastic scattering by a central force ! conservation of
" kinetic energy
" magnitude of the momentum (direction generallychanges)
# total change in momentum,
!p $ p1 %p,
is along the z& axis
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Conservation of Momentum
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Central Force Scattering
• for this geometry,
!p/2M1v
= Sin!!
2
"!" !p = 2M1v Sin
!!2
"
• Newton’s second law:
F = p " dp = Fdt
• the force F is the Coulomb repulsion between the scatteredparticle and the target particle:
F = 14"#0
Z1Z2e2
r2
#0: electric constant; Z1,2: atomic numbers; e: electron charge;r : distance.
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Central Force Scattering
• since the angular momentum is conserved under a centralforce,
M1vb = M1r2d!dt
! dtd!
= r2
vb
• inserting this result and the expression for the Coulombforce into the above equation for !p yields
!p = 14"#0
Z1Z2e2
r2
!Cos[!] r
2
vbd!
= 14"#0
Z1Z2e2
vb
!Cos[!]d!
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Central Force Scattering
• this implies
!p = 14!"0
Z1Z2e2
vb(Sin[#2]! Sin[#1])
• according to the geometry of the scattering process,
#1 = !#0, #2 = +#0, 2#0 + $ = !
• it follows that
Sin[#2]! Sin[#1] = 2 Sin!!
2! $
2
"
• combining this with the above equation for !p yields
!p = 14!"0
Z1Z2e2
vb
#2 Sin
!!2! $
2
"$= 2
4!"0
Z1Z2e2
vbCos
!$2
"
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Central Force Scattering
• this yields a relationship between the impact parameter band the scattering angle !:
b = 14"#0
Z1Z2e2
M1v2 Cot!!
2
"= 1
4"#0
Z1Z2e2
2ECot
!!2
"
• using the previous equation
2"bdb = !$[!] · 2" Sin[!]d!,
the scattering cross-section can be expressed as
$[!] = ! bSin[!]
dbd!
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Central Force Scattering
• inserting the geometrical relationships
Sin[!] = 2 Sin!!
2
"Cos
!!2
"
d Cot!!
2
"= ! d!
2 Sin[!/2]2
yields the cross-section originally derived by Rutherford:
"[!] =#
14#$0
Z1Z2e2
4E
$21
Sin[!/2]4
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Central Force Scattering
• the closest approach d of the scattered particle to the tar-get particle is given by equating its kinetic energy to itspotential energy at a distance d away from the core of thetarget particle:
d = 14!"0
Z1Z2e2
E
• inserting this into the Rutherford expression for the scat-tering cross-section yields
#[$] = (d/4)2
Sin[$/2]4
• for $ = ! this yields
#[!] = (d/4)2
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Central Force Scattering
• for 2 MeV He ions (Z1 = 2 incident on Ag (Z2 = 47) thisyields a closest particle–scatterer distance of
d = 6.8 · 10!5 nm
" much (!) smaller than the Bohr radius(radius of a hydrogen atom, 5 · 10!2 nm)
• the cross-section for scattering to ! = " is
#[!] = 2.98 · 10!10 nm2 = 2.89 · 10!28 m2 # 2.89 barn
1 barn := 10!28 m2
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Conclusion
• conclusions for materials scientists:
The use of an unscreened cross-section is justified.
RBS is a powerful, quantitative method.
• conclusion for physicists and philosophers:
THE WORLD IS “EMPTY!”
